Motion Problems | MathHelp.com - By MathHelp.com
Transcript
| 00:00 | Gordon Rode his bike at 15 mph to go get | |
| 00:04 | his car . He then drove back at 45 mph | |
| 00:10 | . If the entire trip took him eight hours , | |
| 00:13 | how far away was his car ? Let's start by | |
| 00:17 | setting up a diagram just like we did in the | |
| 00:19 | last section so that we can visualize what's going on | |
| 00:24 | first since Gordon rides his bike to go get his | |
| 00:29 | car . Let's start by making an arrow going to | |
| 00:32 | the right to represent the distance that he bikes or | |
| 00:37 | D . B . From there . He picks up | |
| 00:42 | his car and drives back so we make an arrow | |
| 00:46 | going to the left that represents the distance that he | |
| 00:50 | drives or D D . You can see from the | |
| 00:55 | picture that these two distances are equal . So we | |
| 01:00 | can set up an equation that reads the distance that | |
| 01:04 | he bikes or D . B . Equals the distance | |
| 01:07 | that he drives . D . D . Now let's | |
| 01:11 | set up a chart based on our old formula rate | |
| 01:15 | times . Time equals distance . Yeah down the left | |
| 01:30 | side we have biking and driving . Yeah . Now | |
| 01:41 | let's fill out the chart . Gordon's rate biking is | |
| 01:47 | 15 mph . His rate driving is 45 mph . | |
| 01:57 | Be careful with the time column . We know that | |
| 02:00 | his total time is eight hours but we don't know | |
| 02:04 | how long he spent biking and how long he spent | |
| 02:07 | driving . But since the total is eight hours we | |
| 02:11 | can represent the parts as X . & eight -X | |
| 02:19 | . So his distance biking would then be 15 times | |
| 02:22 | x . Or 15 X . In his distance driving | |
| 02:27 | would be 45 times eight minus X . Or 45 | |
| 02:32 | parentheses eight minus X . Remember from our diagram that | |
| 02:39 | the two distances are equal so our equation reads 15 | |
| 02:45 | X Equals 45 times parentheses eight -X . Solving from | |
| 02:55 | here we get X equals six . Now let's look | |
| 03:03 | at what the question is , asking , how far | |
| 03:07 | away was his car , In other words , how | |
| 03:11 | far did he have to bike to get to his | |
| 03:14 | car ? So what was his distance biking ? Well | |
| 03:19 | , if we look at our chart , distance biking | |
| 03:22 | is represented by 15 X , since we know that | |
| 03:29 | X equals six , his distance biking is then 15 | |
| 03:37 | times six or 90 miles , so we know his | |
| 03:43 | car was 90 miles away . |
Summarizer
DESCRIPTION:
This lesson covers mapping diagrams. Students learn that if the x-coordinate is different in each ordered pair in a given relation, then the relation is a function. Students also learn to use mapping diagrams and the vertical line test to determine if a relation is a function.
OVERVIEW:
Motion Problems | MathHelp.com is a free educational video by MathHelp.com.
This page not only allows students and teachers view Motion Problems | MathHelp.com videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.
GRADES:
STANDARDS:
